How do you build a ARD circuit 1

[dawg2]

Hello I have an opt 61c and my vendor is now longer with us and I need to build a new ring down circuit could someone please help me with what loads and info I need to program to build a new circuit. Please any and all help will be greatly appreciated.
Thank you in advance

 

[RickyPone]

ld 20
new
500
cls mnl
ftr hot d 4 xxxx

 

[dawg2]

My vendor previous has in a different route and then they are a true circuit. As aa COT line.

 

[trvlr1]

If you have existing circuits like this, they should terminate on a Universal Trunk card in one of your loops. If you can trace one down to find the TN, then print it in LD 20, you will see the basics of how it needs to be built. you build in LD 14, the prompt ATDN is where you enter the extension of the phone you want it to terminate to. You will need a free port on the Universal Trunk card, they only have 8 ports (or less on older cards) so they cross connect to every other pair, not every pair like a digital or analog card.

 

[dawg2]

THANK YOU FOR THIS HELPFUL POST YES THEY ON A UNIVERSAL TRUNK CARD. WHERE WOULD I PROGRAM THE ROUTE I KNOW THAT THEY HAD THERE OWN ROUTE. BUT THIS IS HELPFUL THANK YOU

 

[trvlr1]

If this is to be used for inbound only, you do not have to add it to a route. If you do intend it to be used for outbound too, then go to LD 21. LTM/enter through. This will list out all you route members. Look through it until you find an existing TN you want to "match" for route information. There should be one or more entries for that particular route looking something like the following:
TYPE TLST
TKTP COT
ROUT 6
DES COT
TN 012 0 00 00 MBER 1 TEST
Your routes may not be COT but the principal still applies. Look at the last entry in that block, when building your new line in LD 14, the RTMB (route member) will be the next available number, in this case 6 2
Hope that helps.

 

[dawg2]

Thank you so much this is exactly what I was looking for and I was able to get everything working properly! Very much appreciated!